Today, I explain the proof of Feige’s theorem, hardness of approximation of MAX -COVER. It relies on NP-hardness of -approximating MAX PROJECTION (also known as LABEL COVER) for all , a result due to Ran Raz and following from PCP and parallel repetition.

Recall that this problem has

input : a regular bipartite graph, and for each edge, a projection constraint : .

output: an assigment of labels in and to vertices.

objective: fraction of constraints satisfied.

Feige’s precise theorem is

Theorem 1 There is a Karp reduction of -approximating MAX PROJECTION to -approximating MAX -COVER.

1.2. The proof

The reduction uses the following gadget. Let , , sets

The of this instance is . Indeed, two sets suffice to cover . But only the obvious pairs achieve this. Every other pair has value .

Also, even a cheating solution taking a bounded number of sets does poorly if it is inconsistent. Consistent here means that covering includes a pair . Inconsistent solutions achieve at most .

Proof: Start with a MAX PROJECTION instance , , . We put a copy of the gadget on top of each edge. For , there is a ground set , a copy of . Overall . For each vertex , define sets to be the ‘s. For , the sets cover . For , the ‘s cover .

Note that can be covered only by pairs .

Let . Let us prove completeness: if there is an assignment satisfying -fraction of MAX PROJECTION constraints, then there exist sets covering . Indeed, consider all the for . It covers since it contains and covers with .

Soundness means that if best assignment satisfies of constraints, then any sets cover at most fraction of . By contrapositive. Assume of covering is , i.e. there is a collection of sets with coverage . We shall decode into an assignement , satsfying of constraints.

We do not need to be clever to achieve . Here is some terminology.

Definition 2 For , let and for , . Then define

is consistent in above sens iff , such that . Since , we have in average. and regularity implies that on average over edges. Now we are ready to decode, i.e. make an assignment by picking for a random element of , unless it is empty, in which case pick it arbitrarily. There remains to estimate the score of , which do not do here. Note that is not greedy.

1.3. What should we remember from this proof

To prove -inapproximability for MAX BLAH, it suffices to reduce from MAX PROJECTION using a gadget (instance of MAX BLAH) with special properties.

Every solution of the gadget should suggest at most a constant number of optimal solutions.

Solution which suggest no optimal solutions should have value .

Theorem 3 This methodology of gadgets is a theorem, but only if you reduce from MAX BIJECTION.

Definition 4 MAX BIJECTION is MAX PROJECTION where ‘s are bijections.

Unfortunately, MAX BIJECTION is not a hard problem. Indeed, -approximating MAX BIJECTION is in P.

Subhash Khot’s Unique Games Conjecture states that , such that MAX BIJECTION is -inapproximable.

Then, above technology implies

Theorem 5 Assuming UGC, inapproximability results can be obtained for many problems in a routine manner.

The point is wether the method allows to get tight inapproximability bounds. Answer is yes for certain problems, but it requires sometimes a highly non trivial and mathematical analysis of the various gadget properties above, especially the last one.